Which of the following is the correct structure of L-fructose?

Q: Which of the following is the correct structure of L-fructose?

D- and L-enantiomers are non-superimposable mirror images of each other at all chiral centers. D-fructose has the -OH group at C-5 positioned on the right in its Fischer projection. L-fructose is the exact enantiomer (mirror image) of D-fructose, meaning all chiral centers (C-3, C-4, C-5) have inverted configurations, with the -OH group at C-5 on the left. Structure (3) correctly displays the Fischer projection of L-fructose with C-3 -OH on the right, and C-4, C-5 -OH groups on the left. L-isome

Year	2026	2025	2024	Total
Questions	9	19	8	36

Q28 jee_main_2025_28_jan_evening Carbohydrates and Glycosidic Linkages

Match List-I with List-II

List-I (Saccharides)	List-II (Glycosidic-linkages found)
(A) Sucrose	(I) $α 1-4$ $\alpha\text{ 1-4}$
(B) Maltose	(II) $α 1-4 and α 1-6$
(C) Lactose	(III) $α 1 - β 2$ $\alpha\text{ 1 - }\beta\text{ 2}$
(D) Amylopectin	(IV) $β 1-4$ $\beta\text{ 1-4}$

Choose the correct answer from the options given below :

A. $(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$
B. $(A)-(IV), (B)-(II), (C)-(I), (D)-(III)$
C. $(A)-(II), (B)-(IV), (C)-(III), (D)-(I)$
D. $(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$

Solution

Related Formula

Glycosidic linkages define the connectivity between monosaccharide units in disaccharides and polysaccharides.

Core Logic

Analyzing each saccharide configuration:

Sucrose: Formed by $α$ $\alpha$-D-glucose and $β$ $\beta$-D-fructose via a $α 1 - β 2$ $\alpha 1 - \beta 2$ glycosidic linkage.
Maltose: Composed of two $α$ $\alpha$-D-glucose units connected by a $α 1-4$ $\alpha 1-4$ glycosidic linkage.
Lactose: Composed of $β$ $\beta$-D-galactose and $β$ $\beta$-D-glucose via a $β 1-4$ $\beta 1-4$ glycosidic linkage.
Amylopectin: A branched polymer of glucose with linear $α 1-4$ $\alpha 1-4$ linkages and branching at $α 1-6$ $\alpha 1-6$ positions.

Step 1: Final Mapping

Matching pairs lead to: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Pattern Recognition

Remember shortcuts for common linkages:

Sucrose is a non-reducing sugar involving the anomeric carbons of both units ( $α 1 - β 2$ $\alpha 1 - \beta 2$).
Lactose has a $β$ $\beta$-linkage ( $β 1-4$ $\beta 1-4$).
Amylopectin represents branched starch ( $α 1-4$ $\alpha 1-4$ and $α 1-6$ $\alpha 1-6$).

Chapter Mix

Class 12 Chemistry: Biomolecules

Q32 jee_main_2025_28_jan_evening Hydrolysis of Carbohydrates

Identify correct conversion during acidic hydrolysis from the following: (A) starch gives galactose. (B) cane sugar gives equal amount of glucose and fructose. (C) milk sugar gives glucose and galactose. (D) amylopectin gives glucose and fructose. (E) amylose gives only glucose. Choose the correct answer from the options given below :

A. (C), (D) and (E) only
B. (A), (B) and (C) only
C. (B), (C) and (E) only
D. (B), (C) and (D) only

Solution

Related Formula

Acidic or enzymatic hydrolysis cleaves the glycosidic linkages to yield constituent monosaccharides:

Polysaccharide/Disaccharide H^+ / H₂O Monosaccharides

Core Logic

Evaluating each conversion statement:

(A) Starch $H^+$ $\xrightarrow{H^+}$ only Glucose (not galactose) $arrow$ $\rightarrow$ Incorrect
(B) Cane sugar (Sucrose) $H^+$ $\xrightarrow{H^+}$ $50%$ $50\%$ Glucose + $50%$ $50\%$ Fructose $arrow$ $\rightarrow$ Correct
(C) Milk sugar (Lactose) $H^+$ $\xrightarrow{H^+}$ Glucose + Galactose $arrow$ $\rightarrow$ Correct
(D) Amylopectin $H^+$ $\xrightarrow{H^+}$ only Glucose (not fructose) $arrow$ $\rightarrow$ Incorrect
(E) Amylose $H^+$ $\xrightarrow{H^+}$ only Glucose $arrow$ $\rightarrow$ Correct

Step 1: Finding the Matching Options

Statements (B), (C), and (E) are strictly correct according to carbohydrate biochemistry properties.

Pattern Recognition

Amylose and amylopectin are both structural components of starch, meaning their hydrolysis yields only D-glucose units. Fructose is obtained from sucrose, while galactose comes exclusively from lactose.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q jee_main_2025_29_jan_morning Carbohydrates - Polysaccharides

List - I (Carbohydrate)	List - II (Linkage / Source)
(A) Amylose	(I) β−C₁−C₄ plant
(B) Cellulose	(II) α−C₁−C₄, α−C₁−C₆ animal
(C) Glycogen	(III) α−C₁−C₄, α−C₁−C₆ plant
(D) Amylopectin	(IV) α−C₁−C₄ plant

Carbohydrates linkage source list diagram for Q43 - JEE Main 2025 Morning — The structural visual index links common natural polysaccharides to glycosidic configurations.

A. (A)-(III), (B)-(II), (C)-(I), (D)-(IV)
B. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
C. (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
D. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)

Solution

Related Formula

Polysaccharides are defined by specific monomer units bound through distinct $α$ $\alpha$ or $β$ $\beta$ glycosidic linkages.

Core Logic

Reviewing biological carbohydrate structural configurations based on standard biochemical definitions:

(A) Amylose: Linear unbranched chain polymer of glucose connected via $α-C₁-C₄$ $\alpha-\mathrm{C}_1-\mathrm{C}_4$ paths in plants $arrow$ $\rightarrow$ (IV).
(B) Cellulose: Linear polymer containing glucose linkages connected exclusively via $β-C₁-C₄$ $\beta-\mathrm{C}_1-\mathrm{C}_4$ paths in plants $arrow$ $\rightarrow$ (I).
(C) Glycogen: Animal storage polysaccharide with highly branched links $α-C₁-C₄$ $\alpha-\mathrm{C}_1-\mathrm{C}_4$ and $α-C₁-C₆$ $\alpha-\mathrm{C}_1-\mathrm{C}_6$ pathways $arrow$ $\rightarrow$ (II).
(D) Amylopectin: Branched plant starch components showing mixed linear $α-C₁-C₄$ $\alpha-\mathrm{C}_1-\mathrm{C}_4$ and branched $α-C₁-C₆$ $\alpha-\mathrm{C}_1-\mathrm{C}_6$ lines $arrow$ $\rightarrow$ (III).

This maps precisely to option (2).

Pattern Recognition

Cellulose represents the primary standard framework containing $β$ $\beta$-linkages exclusively; standard starch components like amylose utilize $α$ $\alpha$-configurations.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q jee_main_2024_01_february_morning Nucleic Acids

If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is:

A. CATTAGCT
B. TACGAAGT
C. GTACTTAC
D. ATGCGACT

Solution

Core Logic

Adenine (A) base pairs with Thymine (T) with 2 hydrogen bonds, and Cytosine (C) base pairs with Guanine (G) with 3 hydrogen bonds.

For the given DNA strand: $A arrow T$ $A \rightarrow T$ $T arrow A$ $T \rightarrow A$ $G arrow C$ $G \rightarrow C$ $C arrow G$ $C \rightarrow G$ $T arrow A$ $T \rightarrow A$ $T arrow A$ $T \rightarrow A$ $C arrow G$ $C \rightarrow G$ $A arrow T$ $A \rightarrow T$

Step 1: Sequence Matching

Given sequence: A T G C T T C A Complementary: T A C G A A G T

Nucleic Acids diagram for Q61 - JEE Main 2024 Morning

Pattern Recognition

DNA base pairing strictly follows Chargaff's rule: A=T and G≡C. Just swap A with T, and C with G in the exact order.

Chapter Mix

Class 12 Chemistry: Biomolecules

Q62 jee_main_2024_29_january_evening Polymers and Monomers

Match List I with List II:

List I (Bio Polymer)	List II (Monomer)
A. Starch	I. nucleotide
B. Cellulose	II. $α$ $\alpha$-glucose
C. Nucleic acid	III. $β$ $\beta$-glucose
D. Protein	IV. $α$ $\alpha$-amino acid

Choose the correct answer from the options given below :

A. A-II, B-I, C-III, D-IV
B. A-IV, B-II, C-I, D-III
C. A-I, B-III, C-IV, D-II
D. A-II, B-III, C-I, D-IV

Solution

Related Formula

Factual knowledge of biopolymers and their fundamental repeating units (monomers).

Core Logic

Analyzing each polymer component:

Starch is a polymer composed entirely of $α$ $\alpha$-glucose units.
Cellulose is a linear structural polymer consisting of linear chains of $β$ $\beta$-glucose units.
Nucleic acids (DNA/RNA) are long chains composed of repeating nucleotide units.
Proteins are polypeptides made from combined $α$ $\alpha$-amino acid sequences.

Step 1: Final Match Alignment

Matching structural links properly leads cleanly to the configuration: A-II, B-III, C-I, D-IV.

Pattern Recognition

Standard memorization trick: Plants store starch using alpha linkers, but build rigid cell walls via beta linkers.

Chapter Mix

Class 12 Chemistry: Biomolecules

Which of the following is the correct structure of L-fructose?

Solution & Explanation

Related Formula

Core Logic

Step 1: Structural Verification

Pattern Recognition

Chapter Mix

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Final Mapping

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Matching Options

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Sequence Matching

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Match Alignment

Pattern Recognition

Chapter Mix

More Biomolecules Questions — jee_main_2025_03_april_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back